I am trying to solve following puzzle:
We are given functions $f$, where $f(x) > 0$ and $F := \int_0^x f(t) dt$ and some real $p>1$.
Does $\int_0^\infty f(x)^p e^{-x}dx < \infty$ imply $\int_0^\infty F(x)^pe^{-x}dx < \infty$ ?
This inequality looks a bit similiar to Hardy's, but now I begin thinking that it is too strong and just not true. Help greatly appreciated.
Either I have solved it or I am stupid. Second is very possible. Please verify.
First, let us notice that by Jensen we obtain $$ \left(\int_0^\infty f(t) dt \right)^p \leq \int_0^\infty f(t)^p dt $$
Now let us write $$ \int_0^\infty F(x)^p e^{-x} dx = \int_0^\infty \left(\int_0^x f(t) dt \right)^p e^{-x} dx \leq \int_0^\infty e^{-x} \left(\int_0^x f(t)^p dt \right) dx, $$
where the last inequality is application of above stated Jensen's.
Now, by Fubini's we may write $$ \int_0^\infty e^{-x} \left(\int_0^x f(t)^p dt \right) dx = \int_0^\infty f(t)^p \left(\int_t^\infty e^{-x} dx \right) dt = \int_0^\infty f(t)^p e^{-t} dt, $$
where the last equality is obtained by integrating $\int_t^\infty e^{-x} = e^{-t}$.
Help with verifying or disproving above is again, appreciated. Thank you for your time.